\(a.\left(x+5\right)^2-36=0\)
➜\(\left(x-5\right)^2-6^2=0\)
➜\(\left(x-5-6\right)\left(x-5+6\right)=0\)
➜\(\left[{}\begin{matrix}x-5-6=0\\x-5+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)
Vậy T/n của PT là \(S=\left\{11;-1\right\}\)
b.\(1+\frac{x-1}{3}=\frac{2x+1}{6}-2\)
➜ \(\frac{6}{6}+\frac{2\left(x-1\right)}{6}=\frac{2x+1}{6}-\frac{12}{6}\)
➜\(6+2x-2=2x+1-12\)
➜\(2x-2x=-12+1-6\)
➜\(0x=-12\)
➜ \(x\) vô nghiệm
c.\(\left|2x-1\right|=x-1\)
➜\(\left[{}\begin{matrix}2x-1=x-1\\2x-1=-x+1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{2}{3}\end{matrix}\right.\)
d.\(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{4-6x}{x^2-4}\)
\(ĐKXĐ:x\ne2\) và \(x\ne-2\)
➜\(\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{4-6x}{\left(x-2\right)\left(x+2\right)}\)
⇔\(\left(x-1\right)\left(x-2\right)-x\left(x+2\right)=4-6x\)
⇔\(x^2-2x-x+2-x^2-2x=4-6x\)
⇔\(-5x+6x=4-2\)
⇔\(x=2\)
Vậy.........................
# Học tốt #
